Theorem freq2 4472

Description: Equality theorem for the well-founded predicate. (Contributed by NM, 3-Apr-1994.)

Assertion

Ref	Expression
freq2	⊢ (𝐴 = 𝐵 → (𝑅 Fr 𝐴 ↔ 𝑅 Fr 𝐵))

Proof of Theorem freq2

Dummy variable 𝑠 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		frforeq2 4471	. . 3 ⊢ (𝐴 = 𝐵 → ( FrFor 𝑅𝐴𝑠 ↔ FrFor 𝑅𝐵𝑠))
2	1	albidv 1873	. 2 ⊢ (𝐴 = 𝐵 → (∀𝑠 FrFor 𝑅𝐴𝑠 ↔ ∀𝑠 FrFor 𝑅𝐵𝑠))
3		df-frind 4458	. 2 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑠 FrFor 𝑅𝐴𝑠)
4		df-frind 4458	. 2 ⊢ (𝑅 Fr 𝐵 ↔ ∀𝑠 FrFor 𝑅𝐵𝑠)
5	2, 3, 4	3bitr4g 223	1 ⊢ (𝐴 = 𝐵 → (𝑅 Fr 𝐴 ↔ 𝑅 Fr 𝐵))

Syntax hints:   → wi 4   ↔ wb 105  ∀wal 1396   = wceq 1398   FrFor wfrfor 4453   Fr wfr 4454

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2216

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1812  df-clab 2221  df-cleq 2227  df-clel 2230  df-nfc 2375  df-ral 2527  df-in 3220  df-ss 3227  df-frfor 4457  df-frind 4458

This theorem is referenced by:  weeq2  4483